Question

\[xy({{x}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zx({{z}^{2}}-{{x}^{2}})\] can be represented symbolically as?
(a) \[\sum{xyz({{x}^{2}}-{{y}^{2}})}\]
(b) \[\sum{xy({{x}^{2}}-{{y}^{2}})}\]
(c) \[\sum{xy(x-y)}\]
(d) \[\sum{xy(xy-z)}\]

Answer 1

Hint: We will first assume \[f(x)=xy({{x}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zx({{z}^{2}}-{{x}^{2}})\] and then we will first substitute in this expression y in place of x and then solve. Similarly we will substitute in this expression z in place of x and solve it. We will use these two results to get our answer.

Complete step-by-step answer:
The expression mentioned in the question is \[xy({{x}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zx({{z}^{2}}-{{x}^{2}})\].
So let \[f(x)=xy({{x}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zx({{z}^{2}}-{{x}^{2}})........(1)\]
Now substituting y in place of x in equation (1) we get,
\[f(y)=yy({{y}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zy({{z}^{2}}-{{y}^{2}})........(2)\]
Now cancelling similar terms in equation (2) and also multiplying the terms in equation (2) by opening the brackets we get,
\[f(y)=0+{{y}^{3}}z-y{{z}^{3}}+{{z}^{3}}y-{{y}^{3}}z........(3)\]
Now we see that all the terms in equation (3) are getting cancelled so we get,
\[f(y)=0........(4)\]
Now substituting z in place of x in equation (1) we get,
\[f(y)=zy({{z}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zz({{z}^{2}}-{{z}^{2}})........(5)\]
Now cancelling similar terms in equation (5) and also multiplying the terms in equation (5) by opening the brackets we get,
\[f(y)={{z}^{3}}y-z{{y}^{3}}+{{y}^{3}}z-y{{z}^{3}}+0........(6)\]
Now we see that all the terms in equation (6) are getting cancelled so we get,
\[f(z)=0........(7)\]
So from equation (4) and equation (7) we can say that, \[\sum{xy({{x}^{2}}-{{y}^{2}})}=xy({{x}^{2}}-{{y}^{2}})+yz({{y}^{2}}-{{z}^{2}})+zx({{z}^{2}}-{{x}^{2}})\]. Hence the correct answer is option (b).

Note: Remembering the concept of the expansion of the polynomials and multiplication of variables is the key here. We in a hurry can make a mistake in solving equation (2) and equation (5) as we can write some other power in place of actual powers or x in place of y or z in place of y.